REFRIGERATION CYCLES 761
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\M-therm\Th14-4.pm5
(iii)Theoretical piston displacement per minute
= Mass flow/min. × specific volume at suction
= 18.3 × 0.07874 = 1.441 m^3 /min.
(iv)Theoretical power required to run the compressor
= Mass flow of refrigerant per sec. × compressor work/kg=
18 3
60.
× (h 3 – h 2 ) =18 3
60.
(210.6 – 188.33) kJ/s = 6.79 kJ/s
or 6.79 kW. (Ans.)
(v)Heat removed through the condenser per min.
= Mass flow of refrigerant × heat removed per kg of refrigerant
= 18.3 (h 3 – hf 4 ′) = 18.3 (210.6 – 57.19) = 2807.4 kJ/min. (Ans.)
(vi)Theoretical bore (d) and stroke (l) :
Theoretical piston displacement per cylinder
=Total displacement per minute
Number of cylinder
=
1 441
2.
= 0.7205 m^3 /min.
Also, length of stroke = 1.25 × diameter of piston
Hence, 0.7205 = π/4 d^2 × (1.25 d) × 1000
i.e., d = 0.09 m or 90 mm. (Ans.)
and l = 1.25 d = 1.25 × 90 = 112.5 mm. (Ans.)
+Example 14.26. A refrigeration system of 10.5 tonnes capacity at an evaporator tem-
perature of – 12°C and a condenser temperature of 27°C is needed in a food storage locker. The
refrigerant ammonia is sub-cooled by 6°C before entering the expansion valve. The vapour is 0.95
dry as it leaves the evaporator coil. The compression in the compressor is of adiabatic type.
Using p-h chart find :
(i)Condition of volume at outlet of the compressor
(ii)Condition of vapour at entrance to evaporator
(iii)C.O.P.
(iv)Power required, in kW.
Neglect valve throttling and clearance effect.
Solution. Refer Fig. 14.36.
Using p-h chart for ammonia,
- Locate point ‘2’ where – 12°C cuts 0.95 dryness fraction line.
- From point ‘2’ move along constant entropy line and locate point ‘3’ where it cuts
constant pressure line corresponding to + 27°C temperature. - From point ‘3’ follow constant pressure line till it cuts + 21°C temperature line to get
point ‘4’. - From point ‘4’ drop a vertical line to cut constant pressure line corresponding to – 12°C
and get the point ‘5’.
The values as read from the chart are :
h 2 = 1597 kJ/kg
h 3 = 1790 kJ/kg
h 4 = h 1 = 513 kJ/kg
t 3 = 58°C
x 1 = 0.13.